$\newcommand{\Z}{\mathbb{Z}}$
Let $A = \Z[[x]]$ be the ring of power series.
We know that this is not a PID.
In particular a submodule of the free $A$-module $\bigoplus_{i=1}^nA$ does not have to be free by general theory.
I wonder if there is some improvement given our specific case (i.e. formal series) more precisely:
- What are examples of non free submodules of $\bigoplus_{i=1}^nA$?
- If $\bigoplus_{i=1}^nA = M_1 \oplus M_2 $ where $M_1$ and $M_2$ are submodules of $\bigoplus_{i=1}^nA$, can we conclude that $M_i$ is free?
I have seen examples e.g. $A = \Z/6\Z$ that splits as two non free submodules $\Z/2\Z\oplus \Z/3\Z$ , but I am curious to know if this happens also when $A = \Z[[x]]$.
My idea to prove 2:
$\newcommand{\R}{\mathbb {Q}}$
$\newcommand{\sumn}[1]{\bigoplus_{i=1}^{#1}}$
By tensoring with $\R$, we obtain that $$\sumn{n} (\mathbb{Z}[[x]]\otimes_{\mathbb{Z}}\R)\simeq (M_1\otimes_{\mathbb{Z}} \R) \oplus (M_2\otimes_{\mathbb{Z}} \R)$$
as $\R[[x]]$-modules. Since $\R[[x]]$ is a PID, $$M_i\otimes_{\mathbb{Z}} \R\simeq \sumn{n_1} \R[[x]]$$ with $n_1+n_2 = n.$
Can we conclude from this that $M_i\simeq \sumn{n_1}\Z[[x]]$?
Best Answer
As you were told in the comments, the ideal $(2,X)$ is not free.
From this answer it follows that every finitely generated projective $\mathbb Z[[X]]$-module is extended, that is, if $P$ is a finitely generated projective $\mathbb Z[[X]]$-module there is a finitely generated projective $\mathbb Z$-module $Q$ such that $P\simeq Q\otimes_{\mathbb Z}\mathbb Z[[X]]$. But $Q$ is free, so $P$ is also free.